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A heat transfer of 9.0 times 10^5 J is required to

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Answer #1

The required equation is Q =m*Cp*dT

Latent heat of fusion of ice = 333.5 kJ/kg
Specific heat capacity, ice: 2.108 kJ /kg.
and that of water is 4.186 kJ /kg

first you must elevate the block of m grams from -15 to 0 C
so Q1= m*2.108*(0-(-15))= m*2.108*15

after you melt the ice Q2= m*333.35

after you rise the temperature from 0 to 10oC Q3= m*4.187*10

Total heat Q = Q1 + Q2 + Q3

9.0*102 kJ = m (2.108*15 + 333.35 + 4.187*10)kJ

m = 9.0*102 J / 406.84 = 2.21 kg

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